化工原理下册第三版陈敏恒习题解答.docx

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1、=1.95x10-3=7.28xW43、（D-f=0.1101.34.0106=249510=-ISrn第八章气体吸收1、解：查30。C水的Ps=424KpaP=P-PS=IoI.3-4.24=97.06KPa一=2=&.=起_=卫_=97.06厚XIO-3=1879MXeCeCeMCSMS2.857102182、解：查25。，8厂出。系统=1661x1SV/，设当地大气压为Iatm即1.033at,且不计溶剂分压。P1=10+1,033=11.033aZ=1.08x10w2（绝）P2=0.2+1.033=1.233=1.21102W（绝）：p=EXrp=FQy-y1-1.0810解：20。C

2、时，5=4.0610=4.06X1O60.311.661105y2-1.211021.0-F-1.661105对稀溶液，其比质量分率X=Me-19510-3X1=44=4.7810-3CO22O1C72R13X*=44=1.78103CO2O1芯解得=1.38x10-3w=1.3810-832103=44.27wgw3（2）tn=17.7wgw34、解：为0Cb系统，20。C时，E1=0.53710Wa=0.53705MPa-f=o=-Ex0.5371050.810-5101.3=4.2410(勺-X)=I.09x10=50X=25y,=W2X=25210=50x101=竺=1oXIO_325

3、O-M)2=0.025-0.005=0.02。=QO-2)x10Y=8x10”vO-)2O)i0.020.015=1.33倍=2.67倍(xe-x)2_8.010(Xe-X)I3.010-46、解：查20。C水，pja=2.3346KPat=101.3-2.3346=98.9654P32=101.3-1.33=99.97OCZ)o=0.220XW4w2s20CD=0.252104w2s“DP.Pb2D.P.、NA-In=(-RT1RTP.NAAr=2ABMr=2137s=0.59Ar7、解：查300C,D=O.268cw2sfp=995.7kg/w3,饱和蒸汽压a=31.82阳wPBI=760

4、-31.82=728=IAAmfnHgp12=760fnfnHgPA17601nIn州728二=迎=1.02外744纥d=处，b=001+/MM=念凡-加W=MDPf0(0.01+力)劭=1(p,0H+-H2)MDp1P2号需鬻04+Rw)=.44(fs=6.7day51HPH101.34.62”小8、解：(1)=+=+=+=0.542Kakak1akyak1a64.616.6Kya=PKGa=549KjnOiKh)HoG-G16=0.29mKya54.99、解：(1)-=+K=221KrnoIf(SW)KykykX,M=:e8-”)=2.2x1(*x(005-2x001)=6.6XIOY%H

5、(s)降膜塔勺=0.023(也)083(g)0,：p0P,Ds上=kCC1nk=上产=号与P无关PbPySyEpm=m=W=I.25PP-=-+-,2.8110K1kyHyy-y,e=0.05-.25Q.0VK2-心=1.05x10N广纺=577%NA10、证明：.-1=1_+二Ky匕kxKvakva1akYZZ即HQG=挺1C：y=mx:.x=mKy而H=HaoG=Hmo1两式相除得Noq=ANo1即Na=工坊G证毕A11、证明：由物料衡算得x=x2+-(y-y2)1低浓度吸收ye=wxmG”=+万一3一乃)1Nog=上=1dyy-y外加G、kCtnGmGz入0-2)Q-)+(丁乃一叩2)1

6、1a1,4mG、fnG1(1一丁1+-；-乃一缈2=.In11-2乃一11(x1-xz)=G(y1-y2)mGmG-,2=wx11aZjG,=-=-=0,033bno1sM29/o=-=0.052=0.265w的aH=HQGNC)G=0.26580,9=2.14w16、解：(1)=0.9=1,3277=1.30.9w=1.17w吧=-1=O.855Z1.17w1.17NCfG=1-1(1-)+=51n(1-0.855)+0.855=5.761-fn11F11-0.8551-0.9TN=Ho%=0.8x5.76=461(2)=099=1,37=1.30.99w=1.29wG=-=-1=0.777

7、11.29加1.29Nog=51n(1-0.777)+0.777=14.0。1-0.7771-0.99H=HoGN0G=0.8x14.0=11.2w(3)比较(1)、(2)可知，当回收率提高，则吸收剂利用量为原来的口竺二1倍1.17w由W=1.筋q可知,吸收剂利用量与回收率成正比。17、解：(1)Z=278bwow2-M18(=上生=丝TH曾=1.76x10，f545x2.5Xi。-、。Gm=(j)ttm0i=176X1Q-3278=0.489?/w227R(2)-=695w=545G0.4此时，塔高不受限制，则可在塔顶达到汽液平衡，贝J:%二痛=5452,510-5=0,0136x=x-y=

8、2.5x10-0.0136=5.431021Zzi69518、解：*)Eia-他哈卜=哈)b=143(刍Eia=%十=*=/+(1FM=2nT%乃0.91=勿-得”0.77mia=E卦=自=1.4哈.=1.43(刍Ein=1437=1.430.7w=1用说明操作线与平衡线斜率相等，即推动力处处相等.由心X2-加叼1-707=t=2331U./H1=H1H=HaGNOG=1.22.33=22m19、解：(I)M廿58再=60/58+1W18=0183G=U=2250x273=920批.打22.4722.4298畀蕊=造低r”861=20.386=0,772A1Noq=-r1n(1-)-+-=-1

9、n(1-0.772)+0.772=7.1911-0.7720.0026Ac4G,4x92.2,G=T=1Vfkmo1ItnhD23.1412C11z7Kya=168bwoZ/w3D=0.0467bno1/w2sHOQ0.659(2)回收丙酮量町=G-乃)=92x(0.05-0.0026)=4.36bno1h=253影/h20s解：W=g=6匕=0.592P101.3内=竺=06郃W=O592即操作线与平衡线平行，此时,G0.03必=AyI=AXz=乃一加=必,G0.03ACHog=0.3wOGKya0.1NaG=27=9.00.3.%=滋=丝也得：72=0.00221、解：(1)%=y1(1-

10、瑜=O.01x(1-0.9)=0.0011四3o.667A13(1-1)2i+1)r_11NQQ-n1-AAy2fnx2工y2-21-0.6671r/1、0.01-20.0002CUCc1n(1-0.667)+0.667=5.380.001-20.0002当为上升时，由于H不变，HoG不变，U.NaG=也不变，即HOGCCIIr/ICCf0.01-20.000355.38=1n(1-0.667)X-+0.6671-0.667乃-2x0.00035得乃=OoOI3001-q013=0.870.01(2)当Xz=OOoO2时,C1x1=(y1-2)+x2=(0.01-0.001)+0.0002=3

11、.210313当W=0.00035时,xj=2(1-ey-2)+x2=1(0.01-0.0013)+0.00035=3.2510313当流率增加一倍时，凡酎=ccSocg因为塔高不变，则H=HogNoqi=Jj1=1=410(0-2=357GN.3.57=1n竺3R=41。-1.05100.8解得其=0.000489第九章液体精储1、解：查安托因方程1og*6.031-1211甲恭IOg耳=6.080-t+220.81345+219.5IOIIIogrf=6.031-=2.3479(1) 1080C:108+220.8收=222.80kPaIogrf=6.080-=1.9731108+219.

12、5=94.00IOII1og招=6.031=2.018481C：81+220.8W=IO4.33kPaIogrf=6.080-1345-=1.604181+219.5片=40.19%4=2.5962.484xax(2) %=Q+/)/2=(2.596+2.371)/2=2.484X0.0880.200.300.3970.489计算y0.1930.3830.5150.6210.704实嗡y0.2120.3700.5000.6180.710X0.5920.700.8030.9030.9301.00计算y0.7830.8530.9100.9590.9791.00实验y0.7890.8530.9140.9570.9791.00由y=求得下表1+(a-1)x1+(2.484-I)X取大庆差0.212-0